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Sorry to post this again, but I am still unable to understand it and need help. Please help.1) Using 3(x-3)(x^2-6x+23)^2 as the answer to differentiating f(x)=(x^2-6x+23)^3/2, which I have been able to do, I need to find the general solution of the differential equation dy/dx= (2/27)*(x-3)*((x^2-6x+23)/(y))^1/2, (y>0) in implicit form. (* means to multiply)2) I need the particular solution of the differential equation in(1) for which Y=2 when X=1. 3)I then need this particular solution in explicit form (please state this). I know this is asking a lot but I am really struggling and close to despair with this. Thank you so much.

  • calculus -

    Please confirm or correct typo:
    Using 3(x-3)(x^2-6x+23)^(1/2) as the answer to differentiating f(x)=(x^2-6x+23)^(3/2)


    dy/dx= (2/27)*(x-3)*((x^2-6x+23)/(y))^1/2
    Separate variables and integrate:

    ∫ y^(1/2) dy = ∫ (2/27)(x-3)(x^2-6x+23)^1/2 dx
    After integration,
    (2/3)y^(3/2) = (2/81)(x^2-6x+23)^(3/2)+C
    which reduces to
    Substitute initital conditions x=1.3, y=2 to solve for
    for which (1.3,2) is a particular solution.

    Check the solution by substituting y into the original differential equation and make sure the equation is satisfied for all values of x whenever y>0.

  • calculus -


    In the differential equation we separate


    Let z=x^2-6x+23 then dz=(x^2-6x+23)'dx

    27sqrt(y)dy=sqrt(z)dz Integrating

    27*y^(3/2)/(3/2)=z^(3/2)/(3/2) + C
    18*y^(3/2)=(2/3)*(x^2-6x+23)^(3/2) + C

    2)If y=2 when x=1

    18*2^(3/2)=(2/3)*18^(3/2) + C
    In left side:
    (2/3)*27*2^(3/2)=18*2^(3/2) => C=0

    The particular solution:


    3) 9y-x^2+6x-23=0

  • calculus typo -

    Sorry, MGraph is right. I misread the data as Y=2 and X=1.3 instead of Y=2 and X=1.

    In fact, 3) belongs to a different sentence. However, the process of solution does not change.

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